an airplane with a speed of 40.4 m/s is climbing upward at an angle of 54 degrees with respect to the horizontal. When the planes altitude is 520 m, the pilot releases a package.
a. calculate the distance along the ground measured from a point directly beneath the point of release, to where the package hits the earth.
b. relative to the ground, determine the angle of the velocity vector of the package just before impact (clockwise from the positive x axis)
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To solve this problem, we need to break down the motion of the airplane and the package into horizontal and vertical components.
a. First, let's calculate the time it takes for the package to hit the earth after it is released. We can use the vertical component of motion to find the time:
Initial vertical velocity of the package (Vy) = airplane speed * sine(angle)
Vy = 40.4 m/s * sin(54°)
Vy ≈ 40.4 m/s * 0.809 = 32.7 m/s (rounded to the nearest tenth)
Now, we can use the equation for free fall to find the time of flight (t):
Δy = Vy * t + (1/2) * g * t^2
520 m = 32.7 m/s * t + (1/2) * 9.8 m/s^2 * t^2
Simplifying the equation and solving for t will give us the time:
4.9 t^2 + 32.7 t - 520 = 0
Using the quadratic formula, we can find two possible values for t. Since we need the positive value, we discard the negative solution:
t ≈ 6.14 s (rounded to the nearest hundredth)
Now we can calculate the horizontal distance (range) traveled by the package:
Range = airplane speed * cosine(angle) * time of flight
Range = 40.4 m/s * cos(54°) * 6.14 s
Range ≈ 40.4 m/s * 0.587 * 6.14 s ≈ 146.4 m (rounded to the nearest tenth)
Therefore, the distance along the ground measured from a point directly beneath the point of release to where the package hits the earth is approximately 146.4 meters.
b. To determine the angle of the velocity vector of the package just before impact relative to the ground, we can find the horizontal and vertical components of the package's velocity just before it hits the earth.
Horizontal velocity (Vx) = airplane speed * cosine(angle)
Vx = 40.4 m/s * cos(54°)
Vx ≈ 40.4 m/s * 0.587 = 23.7 m/s (rounded to the nearest tenth)
Vertical velocity (Vy) would be the negative of the initial vertical velocity because the package is moving downward:
Vy = -32.7 m/s
Now we can use trigonometry to find the angle of the velocity vector:
Angle = arctan(Vy / Vx)
Angle = arctan(-32.7 m/s / 23.7 m/s)
Angle ≈ -55.6° (rounded to the nearest tenth)
Clockwise from the positive x-axis, the angle of the velocity vector of the package just before impact relative to the ground is approximately -55.6 degrees.